question_answer

ABCD is a trapezium whose side AD is parallel to BC. Diagonals AC and 80 intersect at O. If \[AO=3,\] \[CO=x-3,\] \[BO=3x-19\] and \[DO=x-5,\] the value of x will be

A)  8, 9     

B)         7, 6

C)  12, 6   

D)  7, 10

Correct Answer: A

Solution :

In any quadrilateral ABCD, when diagonal intersect at O, then \[\frac{AO}{OC}=\frac{OD}{OB}\] \[\frac{AO}{OC}=\frac{OD}{OB}\Rightarrow \frac{3}{x-2}=\frac{x-5}{3x-19}\] \[\Rightarrow \]\[3(3x-19)=(x-3)(x-5)\] \[\Rightarrow \]\[9x-57={{x}^{2}}-8x+15\] \[\Rightarrow \]\[{{x}^{2}}-17x+72=0\] On solving,\[(x-8)(x-9)=0\,\,\,\,\Rightarrow \,\,x=8,9\]